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Chem 14A

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Wave-particle Duality. Heisenberg Uncertainty Principle. Schr dinger Equation ... Wave-Particle Duality ... Wave-Particle Duality. Electron wavelength: ?=h/mv ... – PowerPoint PPT presentation

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Title: Chem 14A


1
Chem 14A
  • Friday, June 29, 2007

2
Lecture Outline
  • Brief review of quantum theory experiments
  • Photoelectric effect
  • Wave-particle Duality
  • Heisenberg Uncertainty Principle
  • Schrödinger Equation
  • Atomic Spectra and Energy

3
Early Quantum Mechanics
  • Max Planck postulated that energy is quantized
    based on blackbody radiation experiments where
    continuous energy absorption was not seen
  • Einstein used this information to develop a
    theory that light is also quantized, or delivered
    in photons, particle-like energy packets with
    energy value Ehn

4
Early Quantum Mechanics
  • Millikan performed a series of experiments to
    prove Einsteins theory that light is quantized
    by using the photoelectric effect

5
Photoelectric Effect
  • A Metal surface is illuminated with incident
    radiation, exciting electrons which are ejected
    from the surface with kinetic energy

6
Photoelectric Effect
7
Photoelectric Effect
  • Electrons were only ejected under certain
    conditions
  • Radiation frequency must be above a certain level
    (threshold value)
  • Electrons ejected immediately without regard to
    radiation intensity (number of photons)
  • Kinetic energy of ejected electrons increases
    linearly with frequency of incident radiation

8
Photoelectric Effect
  • What do the results mean?
  • If light is wave-like, the energy contained in
    one of those waves should depend only on its
    amplitude--that is, on the intensity of the
    light. Lower intensity light would give fewer
    electrons ejected more slowly.
  • If light is particulate, changing the frequency
    will result in different energy values for the
    electrons, but a different intensity will have no
    effect, since each particle has the same energy
    value

9
Photoelectric Effect
  • As a result of the photoelectric effect,
    Einsteins theory that electromagnetic radiation
    is quantized was validated, and he was awarded
    the Nobel prize.
  • Light now has a particulate nature!

10
Work Function
  • Photons of light collide with the metal surface
    electrons, ejecting them
  • Energy required to eject en electron from the
    surface work function (F)
  • Energy of the colliding photon Ehn

11
Work Function
  • The work function describes the difference in
    energy between the photon itself and the energy
    needed to eject the electron
  • Elt F
  • If the photons energy is smaller than the energy
    needed to eject the electron, the electron will
    never be ejected, regardless of the light
    intensity (number of photons)

12
Work Function
  • EgtF
  • If the photons energy is greater than the energy
    needed to eject an electron, the electron is
    ejected with positive kinetic energy KE1/2mv2
  • KE E- F

13
Work Function
  • KE E- F
  • Substituting into the equation gives
  • 1/2mv2hn- F

14
Quick Exercise I.
  • It takes 7.21x10-19J of energy to remove an
    electron from an iron atom. What is the maximum
    wavelength of light that can accomplish this?
  • 1/2mv2hn- F
  • ?c/n
  • c 3.00x108 m/s

15
Answers to Quick Exercise I.
  • 1/2mv2hn- F
  • We want to find the smallest possible frequency
    in order to get the largest possible wavelength,
    since ?c/n

16
Answers to Quick Exercise I.
  • Re-writing the equation in terms of frequency,
    and solving for the lowest possible frequency
    value by setting the velocity to zero

1/2mv2 F
1/2m(0) 7.21x10-19 J
n
n
6.626x10-34 Js
h
h
17
Answers to Quick Exercise I.
  • Solve for the lowest frequency value
  • Then plug into equation for wavelength

n 7.21x10-19J
3.00x108 m/s
1.088x1015 s-1
?
1.088x1015 s-1
6.626x10-34 Js
18
Answers to Quick Exercise I.
  • Wavelength 2.76x10-7 m or 276 nm

19
Wave-Particle Duality
  • Electromagnetic Radiation also has evidence of
    wave-like properties
  • Diffraction Light will show a diffraction
    pattern when passed through a diffraction grating

20
Wave-Particle Duality
  • What is diffraction?
  • Diffraction is a property unique to waves, based
    on the addition and subtraction of several wave
    forms.
  • Constructive Interference when waves are added
    together the amplitude increases
  • Destructive Interference when waves are
    subtracted from each other the amplitude decreases

21
Wave-Particle Duality
combinedwaveform
22
Wave-Particle Duality
  • The collection of high and low intensities
    formed from wave constructive and destructive
    interference produces a pattern of lines behind
    the diffraction grating

23
Wave-Particle Duality
  • Duality applies to electromagnetic radiation as
    seen through experiments, what about matter?
  • DeBroglie proposed that all particles have
    wave-like properties too
  • The DeBroglie Equation ?h/mv
  • mvp linear momentum

24
Wave-Particle Duality
  • Wave-like properties of matter are strongly mass
    dependent
  • Using the DeBroglie Equation
  • Baseball wavelength
  • ?h/mv
  • 6.626x10-34/(.1kg)(35m/s)1.9X10-34 m

25
Wave-Particle Duality
  • Electron wavelength
  • ?h/mv
  • 6.626x10-34/(9.1x10-31kg)(1x107m/s)
  • 7.3X10-11 m
  • It is hard to observe wave properties of
    macroscopic objects, but they do exist

26
Quick Exercise II.
  • Calculate the de Broglie wavelength for
  • A proton with a velocity 5 of the speed of light
  • An electron with a velocity 15 of the speed of
    light
  • c3x108 m/s ?h/mv
  • me 9.1x10-31kg h 6.626x10-34
    Js
  • mp 1.67x10-27kg

27
Answers to Quick Exercise II.
  • 1. 2.6x10-5 nm
  • 2. 1.6x10-2 nm

28
Heisenberg Uncertainty Principle
  • Matter has both wave and particulate properties
  • Waves are delocalized- Their exact location
    cannot be described
  • Particles are localized- Their exact location is
    easily quantifiable

29
Heisenberg Uncertainty Principle
  • Since matter has both wave and particulate
    properties, how do we define its location and
    momentum?

30
Heisenberg Uncertainty Principle
  • We cannot know both the position and momentum of
    a particle to a high degree of precision at the
    same time
  • The more precisely we know a particles position,
    the less precisely we know its momentum and vice
    versa
  • ?x?ph/4p

31
Heisenberg Uncertainty Principle
  • For large particles, such as a baseball the level
    of uncertainty is small.
  • For small particles, such as an electron, the
    level of uncertainty is large- we dont know the
    exact path an electron travels in around the
    nucleus (orbitals are 90 probability
    representations)

32
Quick Exercise III.
  • The hydrogen atom has a radius on the order of
    0.05 nm. Assuming we know the position of an
    electron to an accuracy of 1 of the radius,
    calculate the uncertainty in the velocity of the
    electron.
  • me9.11x10-31 kg
  • h 6.626x10-34 Js

33
Answer to Quick Exercise III.
  • ?x?ph/4p
  • ?x 0.01x0.05nm 0.0005nm 5x10-13m
  • ?ph/4p?x 1.05x10-22 kg m/s
  • ?pm?v
  • ?v ?p/m(1.05x10-22 kg m/s)/9.11x10-31kg
  • ?v 1.15X108 m/s

34
The Schrödinger Equation
  • Austrian Physicist Erwin Schrödinger adopted the
    use of wave functions to describe the path of a
    particle, since it has wave-like character
  • A wave function ? is a mathematical function of
    the position of a particle in three-dimensional
    space (x,y,z)

35
The Schrödinger Equation
  • A wave function squared describes the probablity
    density the probability of finding a particle
    within a certain space
  • Schrodinger developed an equation for calculating
    wave functions
  • H? E? where HHamiltonian operator

36
The Schrödinger Equation
  • Solutions to the Schrodinger Equation
  • There are many different solutions to the
    Schrödinger equation
  • Each solution to the Schrödinger equation
    represents an orbital
  • For example, There are separate wave functions
    for the H 1s, 2s, and 2p orbitals

37
Atomic Spectra and Energy Levels
  • Atomic spectra give further evidence that
    energy is quantized
  • Emission spectra are produced when light emitted
    by atoms is passed through a prism
  • White light passed through a prism gives a
    continuous spectrum (rainbow)
  • Light emitted by excited atoms (ie hydrogen)
    gives discrete spectral lines

38
Atomic Spectra and Energy Levels
  • Absorption spectra are created by shining light
    through an atomic vapor
  • Absorption spectra exhibit discrete quantities of
    energy being absorbed by an atom and give the
    same spectra lines as emission spectra, only they
    are black against a rainbow background

39
Atomic Spectra and Energy Levels
40
Atomic Spectra and Energy Levels
  • Because an atom can emit only certain wavelengths
    of light as indicated by the discrete spectral
    lines, energy is quantized
  • Atoms lose energy in specific amounts, implying
    that electrons are present only in certain energy
    states
  • During an electronic transition, an electron
    moves from a higher energy level to a lower
    energy level and releases energy in terms of a
    photon

41
Atomic Spectra and Energy Levels
  • Bohr Frequency Condition ?Ehn transition energy
    released as a photon
  • Since each line on the spectrum corresponds to a
    particular electronic transition, the spectrum
    can be used to develop an energy-level diagram

42
Atomic Spectra and Energy Levels
  • Rydberg constant all spectral lines
  • ?R(1/n12 1/n22) n1 1,2 n2 n11, n12
  • RRydberg constant3.29x1015 Hz
  • Balmer series visible spectral lines
  • n12 and n23,4,
  • Lyman series ultraviolet spectral lines
  • n11 and n22,3,

43
Atomic Spectra and Energy Levels
  • Using the Rydberg equation to identify spectral
    lines
  • Calculate the wavelength of radiation emitted by
    a hydrogen atom during an electronic transition
    with n23 and n12
  • What color is the light?

44
Atomic Spectra and Energy Levels
  • nR(1/n12 1/n22) R3.29x1015 Hz
  • nR(1/22 1/32) 4.57x1014 Hz
  • ?c/n (3x108m/s)/(4.57x1014 Hz)
  • ?6.57x10-7m or 657nm red light

45
Quick Exercise IV.
  • Calculate the wavelength for a transition between
    n25 and n11
  • What region of the electromagnetic spectrum is
    the light found in ?
  • nR(1/n12 1/n22) R3.29x1015 Hz
  • ?c/n

46
Answers to Quick Exercise IV.
  • Wavelength 94.9nm Lyman Series
  • Ultraviolet region
  • nR(1/12 1/52) 3.158x1015 Hz
  • ?c/n (3x108m/s)/(3.158x1015 Hz)
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